Evaluate the integral: I = integral x cube root of (x-2)
Evaluate the integral: $$ I = \int x \sqrt[3]{x-2} \, dx. $$
Explanation of Method
When $x$ multiplies a root of a linear expression, substitute $u$ for the inside of the root and also solve for $x$ in terms of $u$ so the whole integrand becomes a polynomial-like expression in powers of $u$. Expanding the product then reduces the problem to ordinary power-rule integration.
- Substitute $u$ for the linear expression inside the root.
- Rewrite $x$ in terms of $u$ and expand the product.
- Integrate each resulting power of $u$.
- Substitute back and add the constant of integration.
Step 1 · Use substitution
To simplify this integral, we use the substitution method. Let:
$$ u = x - 2. $$
Then, the derivative of $u$ with respect to $x$ is:
$$ \frac{du}{dx} = 1 \quad \Rightarrow \quad du = dx. $$
Now, solve for $x$ in terms of $u$:
$$ x = u + 2. $$
Substitute into the integral:
$$ I = \int (u+2)\sqrt[3]{u}\,du. $$
This can be written as:
$$ I = \int (u+2)u^{1/3}\,du. $$
Step 2 · Expand the integrand
Now expand the integrand:
$$ I = \int u^{1+1/3} + 2u^{1/3}\,du = \int u^{4/3} + 2u^{1/3}\,du. $$
Step 3 · Integrate each term
First term:
$$ \int u^{4/3}\,du = \frac{u^{4/3+1}}{4/3+1} = \frac{u^{7/3}}{7/3} = \frac{3}{7}u^{7/3}. $$
Second term:
$$ \int u^{1/3}\,du = \frac{u^{1/3+1}}{1/3+1} = \frac{u^{4/3}}{4/3} = \frac{3}{4}u^{4/3}. $$
Multiply by $2$:
$$ 2\int u^{1/3}\,du = 2\cdot \frac{3}{4}u^{4/3} = \frac{3}{2}u^{4/3}. $$
Step 4 · Substitute back $u = x - 2$
Substitute $u = x-2$ back into the results:
$$ I = \frac{3}{7}(x-2)^{7/3} + \frac{3}{2}(x-2)^{4/3} + C. $$
Summary Table
| Term | Antiderivative |
|---|---|
| $u^{4/3}$ | $\tfrac{3}{7}u^{7/3}$ |
| $2u^{1/3}$ | $\tfrac{3}{2}u^{4/3}$ |
Additional Notes
- The integrand is defined for all real $x$, since cube roots (odd roots) are defined for both positive and negative arguments.
- Rewriting $x = u+2$ and expanding $(u+2)u^{1/3}$ into $u^{4/3}+2u^{1/3}$ is the essential trick whenever the outer factor $x$ cannot be absorbed directly into the substitution.
- The result can be checked by differentiating $\tfrac37(x-2)^{7/3}+\tfrac32(x-2)^{4/3}$ with the product and chain rules and confirming it returns $x\sqrt[3]{x-2}$.
Result
The solution to the integral is: